Problem: There is a railroad in Takahashi Kingdom. The railroad consists of N sections, numbered 1, 2, ..., N, and N+1 stations, numbered 0, 1, ..., N. Section i directly connects the stations i-1 and i. A train takes exactly A_i minutes to run through section i, regardless of direction. Each of the N sections is either single-tracked over the whole length, or double-tracked over the whole length. If B_i = 1, section i is single-tracked; if B_i = 2, section i is double-tracked. Two trains running in opposite directions can cross each other on a double-tracked section, but not on a single-tracked section. Trains can also cross each other at a station.

Snuke is creating the timetable for this railroad. In this timetable, the trains on the railroad run every K minutes, as shown in the following figure. Here, bold lines represent the positions of trains running on the railroad. (See Sample 1 for clarification.)

<image>

When creating such a timetable, find the minimum sum of the amount of time required for a train to depart station 0 and reach station N, and the amount of time required for a train to depart station N and reach station 0. It can be proved that, if there exists a timetable satisfying the conditions in this problem, this minimum sum is always an integer.

Formally, the times at which trains arrive and depart must satisfy the following:

* Each train either departs station 0 and is bound for station N, or departs station N and is bound for station 0.
* Each train takes exactly A_i minutes to run through section i. For example, if a train bound for station N departs station i-1 at time t, the train arrives at station i exactly at time t+A_i.
* Assume that a train bound for station N arrives at a station at time s, and departs the station at time t. Then, the next train bound for station N arrives at the station at time s+K, and departs the station at time t+K. Additionally, the previous train bound for station N arrives at the station at time s-K, and departs the station at time t-K. This must also be true for trains bound for station 0.
* Trains running in opposite directions must not be running on the same single-tracked section (except the stations at both ends) at the same time.

Constraints

* 1 \leq N \leq 100000
* 1 \leq K \leq 10^9
* 1 \leq A_i \leq 10^9
* A_i is an integer.
* B_i is either 1 or 2.

Input

The input is given from Standard Input in the following format:


N K
A_1 B_1
A_2 B_2
:
A_N B_N


Output

Print an integer representing the minimum sum of the amount of time required for a train to depart station 0 and reach station N, and the amount of time required for a train to depart station N and reach station 0. If it is impossible to create a timetable satisfying the conditions, print -1 instead.

Examples

Input

3 10
4 1
3 1
4 1


Output

26


Input

1 10
10 1


Output

-1


Input

6 4
1 1
1 1
1 1
1 1
1 1
1 1


Output

12


Input

20 987654321
129662684 2
162021979 1
458437539 1
319670097 2
202863355 1
112218745 1
348732033 1
323036578 1
382398703 1
55854389 1
283445191 1
151300613 1
693338042 2
191178308 2
386707193 1
204580036 1
335134457 1
122253639 1
824646518 2
902554792 2


Output

14829091348

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;

#define LL long long

const int maxN = 2e5 + 10;
const LL inf = 1e18;

struct Segment
{
	LL mn;
	bool lazy;
}tree[maxN * 4 + 1];

int n, K, cnt;
int a[maxN + 1], b[maxN + 1];
int L[maxN + 1], R[maxN + 1], tmp[maxN + 1];
LL sum[maxN + 1], ans;

inline int read()
{
	int num = 0, f = 1;
	char ch = getchar();
	while( !isdigit( ch ) ) { if(ch == '-') f = -1; ch = getchar(); }
	while( isdigit( ch ) ) num = (num << 3) + (num << 1) + (ch ^ 48), ch = getchar();
	return num * f;
}

inline void pushup(int node)
{
	tree[node].mn = min(tree[node << 1].mn, tree[node << 1 | 1].mn);
}

inline void pushdown(int node)
{
	tree[node << 1].mn = tree[node << 1 | 1].mn = inf;
	tree[node << 1].lazy = tree[node << 1 | 1].lazy = true;
	tree[node].lazy = false;
}

inline void build(int node, int l, int r)
{
	tree[node].lazy = false;
	if(l == r) { tree[node].mn = -tmp[l]; return; }
	int mid = (l + r) >> 1;
	build(node << 1, l, mid);
	build(node << 1 | 1, mid + 1, r);
	pushup(node);
}

inline void modify(int node, int l, int r, int x, int y)
{
	if(x > y) return;
	if(x <= l && r <= y)
	{
		tree[node].mn = inf;
		tree[node].lazy = true;
		return;
	}
	if(tree[node].lazy) pushdown(node);
	int mid = (l + r) >> 1;
	if(x <= mid) modify(node << 1, l, mid, x, y);
	if(y > mid) modify(node << 1 | 1, mid + 1, r, x, y);
	pushup(node);
}

inline void change(int node, int l, int r, int x, LL num)
{
	if(l == r) { tree[node].mn = min(tree[node].mn, num); return; }
	if(tree[node].lazy) pushdown(node);
	int mid = (l + r) >> 1;
	if(x <= mid) change(node << 1, l, mid, x, num);
	else change(node << 1 | 1, mid + 1, r, x, num);
	pushup(node);
}

inline LL query(int node, int l, int r, int x, int y)
{
	if(x > y) return inf;
	if(x <= l && r <= y) return tree[node].mn;
	if(tree[node].lazy) pushdown(node);
	int mid = (l + r) >> 1; LL ans = inf;
	if(x <= mid) ans = min(ans, query(node << 1, l, mid, x, y));
	if(y > mid) ans = min(ans, query(node << 1 | 1, mid + 1, r, x, y));
	return ans;
}

inline void get(int node, int l, int r)
{
	if(l == r) { ans = min(ans, tree[node].mn + tmp[l]); return; }
	if(tree[node].lazy) pushdown(node);
	int mid = (l + r) >> 1;
	get(node << 1, l, mid);
	get(node << 1 | 1, mid + 1, r);
}

int main()
{
	n = read(), K = read();
	for(int i = 1; i <= n; i++)
	{
		a[i] = read(), b[i] = read();
		sum[i] = sum[i - 1] + a[i];
		if(b[i] == 1) L[i] = (K - 2ll * sum[i - 1] % K) % K, R[i] = (K - 2ll * sum[i] % K) % K;
		else L[i] = 0, R[i] = K - 1;
		if(b[i] == 1 && 2ll * a[i] > K)
		{
			puts("-1");
			return 0;
		}
		tmp[++ cnt] = L[i]; tmp[++ cnt] = R[i];
	}

	sort(tmp + 1, tmp + cnt + 1);
	cnt = unique(tmp + 1, tmp + cnt + 1) - tmp - 1;

	build(1, 1, cnt);

	for(int i = 1; i <= n; i++)
	{
		if(b[i] == 2) continue;
		int l = lower_bound(tmp + 1, tmp + cnt + 1, L[i]) - tmp, r = lower_bound(tmp + 1, tmp + cnt + 1, R[i]) - tmp;
		if(l <= r)
		{
			LL t = min(query(1, 1, cnt, 1, l - 1) + tmp[l], query(1, 1, cnt, r + 1, cnt) + K + tmp[l]);
			modify(1, 1, cnt, 1, l - 1); modify(1, 1, cnt, r + 1, cnt);
			change(1, 1, cnt, l, t - tmp[l]);
		}
		else
		{
			LL t = query(1, 1, cnt, r + 1, l - 1) + tmp[l];
			modify(1, 1, cnt, r + 1, l - 1);
			change(1, 1, cnt, l, t - tmp[l]);
		}
	}

	ans = inf;
	get(1, 1, cnt);

	printf("%lld", ans + 2ll * sum[n]);
	return 0;
}